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Liono4ka [1.6K]

2 years ago

7

You observe three carts moving to the left. Cart A moves to the left at nearly constant speed. Cart B moves to the left, gradual

ly speeding up. Cart C moves to the left, gradually slowing down. Which cart or carts, if any, experience a net force to the left? Cart A Cart C Cart B

1 answer:

Lady bird [3.3K]2 years ago

6 0

Answer:cart B

Explanation:

For cart A speed is constant therefore there is no acceleration because acceleration is rate of change of velocity

thus there is no net force

For cart B there is change in velocity in the left direction , so there is net acceleration towards left

$Force=mass\times acceleration$

so there is net force in the left direction

For cart C there is decrease in velocity i.e. negative acceleration or deceleration . Therefore there is a net force towards right which opposes the motion

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<em /> $m_{1} v_{i1} +m_{2} v_{i2} =m_{1} v_{f1} +m_{2} v_{f2}$ <em />

<em>so since the bottle is at rest firstly, therefore </em> $v_{i2} =0$ <em />

<em /> $m_{1} v_{i1} +m_{2} (0) =m_{1} v_{f1} +m_{2} v_{f2}$ <em />

<em /> $m_{1} v_{i1} =m_{1} v_{f1} +m_{2} v_{f2}$ <em> </em><em>equation 1</em>

so now substitute $v_{f1}$ into equation 1

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<em /> $m_{1} v_{i1} = -0.3m_{1}v_{i1} +m_{2} v_{f2}$ <em />

<em>collect the like terms</em>

$m_{1} v_{i1} +0.3m_{1}v_{i1} =m_{2} v_{f2}$

$1.3m_{1} v_{i1} =m_{2} v_{f2}$

divide both side by $m_{2}$

$v_{f2}=\frac{1.3m_{1} v_{i1}}{m_{2} }$

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<em />

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